Zero group refractive index

  • #1
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Homework Statement

.[/B]
I am attempting to determine the group refractive index of a laser cavity at it's resonance frequency.

Homework Equations

.[/B]
\begin{align*}
\frac{2\omega n L}{c} &= 2m\pi
\end{align*}

\begin{align*}
n_g &= n + \omega \frac{dn}{d\omega}
\end{align*}

3. The attempt at the solution.
I have considered a uniform plane wave propagating within the cavity satisfying the following relation
\begin{align*}
\frac{2\omega n L}{c} &= 2m\pi
\end{align*}
where ω is the angular frequency, n is the effective refractive index and L is the length of the laser cavity. I have derived the group refractive index as follows
\begin{align*}
n_g &= n + \omega \frac{dn}{d\omega} \\
&= \frac{c m \pi}{\omega L} - \frac{c m \pi}{ \omega L}\\
&= 0
\end{align*}

If this is correct, I don't understand what this is physically entailing. Any insight would be much appreciated! Thank you.
 

Answers and Replies

  • #2
You are calculating the group index of refraction for a monochromatic wave, which is of course zero.
 

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